= Solution
An <ideal> is an additive subgroup of a <ring> closed under multiplication by arbitrary ring elements; in a potentially noncommutative ring, a two-sided ideal must absorb multiplication on both sides. For a <ring homomorphism> $\phi$, its <kernel of a ring homomorphism> $K=\{r:\phi(r)=0\}$ contains zero, is closed under subtraction, and satisfies $\phi(rk)=\phi(r)\phi(k)=0$ and $\phi(kr)=0$. Thus $K$ is an ideal.
Define $\widetilde\phi:R/K\to S$ by $r+K\mapsto\phi(r)$. It is well-defined because two representatives differ by an element of $K$, preserves addition and multiplication, is surjective by hypothesis, and is injective because its zero class is precisely $K$. Therefore $\boxed{R/\ker\phi\cong S}$, the <first isomorphism theorem for rings>.
Apply this to the <evaluation homomorphism> $\mathbb C[z]\to\mathbb C$, $p\mapsto p(1)$. Constants prove surjectivity, and division by $z-1$ shows that its <kernel of a ring homomorphism> is $(z-1)$. Hence $\boxed{\mathbb C[z]/(z-1)\cong\mathbb C}$.
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